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In Calculus by Michael Spivak, he defines uniform continuity as follows:

The function $f$ is uniformly continuous on an interval $A$ if for every $\epsilon > 0$ there is some $\delta > 0$ such that, for all $x$ and $y$ in $A$, if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon$.

I understand the difference between continuity and uniform continuity. However, I have been unable to distinguish between the definitions of continuity and uniform continuity. For instance, when reading the aforementioned definition by Spivak and then compared it to definitions of continuity on the internet, I am unable to see/understand the differences.

I would greatly appreciate it if people could please take the time to clarify and elaborate on the differences between the definitions. Please use both definitions to make clear comparisons.

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    $\begingroup$ @ΘΣΦGenSan I already read the answers to that question. What I am asking is different: What are the specific differences between the two definitions -- not the concepts. $\endgroup$ Commented Mar 3, 2017 at 11:04
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    $\begingroup$ Ah okay, ignore my comment. I withdraw the flag for duplicate. $\endgroup$ Commented Mar 3, 2017 at 11:05

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$f$ continuous on $A$ if for all $\varepsilon >0$ and for all $y \in A$ there exists $\delta > 0$ such that $|f(x)-f(y) | \le \varepsilon$ for all $x \in A$ with $|x-y| < \delta$.

$f$ is uniformly continuos on $A$ if for all $\varepsilon > 0$ there exists $\delta >0$ such that $|f(x)-f(y)| \le \varepsilon$ for all $x,y \in A$ with $|x-y|< \delta$.

The difference in the definitions comes in the fact that for continuity first you fix $y \in A$ and then you find $\delta$, while for uniform continuity you find a $\delta$ which works for all $y$. In other words, the $\delta$ in the continuity statement depends on $\varepsilon$ and on the point you want to check continuity, while in the uniform case, $\delta$ depends only on $\varepsilon$ (i.e. it is uniform with respect to $A$).

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    $\begingroup$ Very late here but just wanted to say that this is a perfectly crystal clear answer. Thanks! $\endgroup$
    – EE18
    Commented Jul 9, 2020 at 14:31

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